有一组解恒定不变,表示这组解与a的取值无关
(a-1)x+(a+2)y+5-2a=0
ax - x + ay + 2y + 5 - 2a = 0
提取a得,a(x + y - 2)- x + 2y + 5 = 0
所以x + y - 2 = 0
-x + 2y + 5 = 0
所以x = 3,y = -1
有一组解恒定不变,表示这组解与a的取值无关
(a-1)x+(a+2)y+5-2a=0
ax - x + ay + 2y + 5 - 2a = 0
提取a得,a(x + y - 2)- x + 2y + 5 = 0
所以x + y - 2 = 0
-x + 2y + 5 = 0
所以x = 3,y = -1